The ratio of escape velocity at earth (υe) to the escape velocity at a planet (υp) whose radius and mean density are twice as that of earth is :
Answer & explanation
Correct answer: option 2
\(v_e = \sqrt{\frac{2GM}{R}} \)
M = \(\frac{4}{3} \pi R^3 \rho\)
⇒ \(v_e = \sqrt{\frac{8 \pi G \rho R^2}{3}} \)
⇒ \(v_p = \sqrt{\frac{8 \pi G (2\rho) (2R)^2}{3}} \)
⇒ vp = 2\(\sqrt{2}\) ve
∴ \(\frac{v_e}{v_p}\) = \(\frac{1}{2\sqrt{2}}\)